\(\int x^3 \arccos (a+b x^4) \, dx\) [71]

Optimal result

Integrand size = 12, antiderivative size = 47 \[ \int x^3 \arccos \left (a+b x^4\right ) \, dx=-\frac {\sqrt {1-\left (a+b x^4\right )^2}}{4 b}+\frac {\left (a+b x^4\right ) \arccos \left (a+b x^4\right )}{4 b} \]

[Out]

1/4*(b*x^4+a)*arccos(b*x^4+a)/b-1/4*(1-(b*x^4+a)^2)^(1/2)/b

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6847, 4888, 4716, 267} \[ \int x^3 \arccos \left (a+b x^4\right ) \, dx=\frac {\left (a+b x^4\right ) \arccos \left (a+b x^4\right )}{4 b}-\frac {\sqrt {1-\left (a+b x^4\right )^2}}{4 b} \]

[In]

Int[x^3*ArcCos[a + b*x^4],x]

[Out]

-1/4*Sqrt[1 - (a + b*x^4)^2]/b + ((a + b*x^4)*ArcCos[a + b*x^4])/(4*b)

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 4716

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcCos[c*x])^n, x] + Dist[b*c*n, Int[
x*((a + b*ArcCos[c*x])^(n - 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 4888

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCos[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \arccos (a+b x) \, dx,x,x^4\right ) \\ & = \frac {\text {Subst}\left (\int \arccos (x) \, dx,x,a+b x^4\right )}{4 b} \\ & = \frac {\left (a+b x^4\right ) \arccos \left (a+b x^4\right )}{4 b}+\frac {\text {Subst}\left (\int \frac {x}{\sqrt {1-x^2}} \, dx,x,a+b x^4\right )}{4 b} \\ & = -\frac {\sqrt {1-\left (a+b x^4\right )^2}}{4 b}+\frac {\left (a+b x^4\right ) \arccos \left (a+b x^4\right )}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91 \[ \int x^3 \arccos \left (a+b x^4\right ) \, dx=\frac {-\sqrt {1-\left (a+b x^4\right )^2}+\left (a+b x^4\right ) \arccos \left (a+b x^4\right )}{4 b} \]

[In]

Integrate[x^3*ArcCos[a + b*x^4],x]

[Out]

(-Sqrt[1 - (a + b*x^4)^2] + (a + b*x^4)*ArcCos[a + b*x^4])/(4*b)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {\left (b \,x^{4}+a \right ) \arccos \left (b \,x^{4}+a \right )-\sqrt {1-\left (b \,x^{4}+a \right )^{2}}}{4 b}\)	\(40\)
default	\(\frac {\left (b \,x^{4}+a \right ) \arccos \left (b \,x^{4}+a \right )-\sqrt {1-\left (b \,x^{4}+a \right )^{2}}}{4 b}\)	\(40\)

[In]

int(x^3*arccos(b*x^4+a),x,method=_RETURNVERBOSE)

[Out]

1/4/b*((b*x^4+a)*arccos(b*x^4+a)-(1-(b*x^4+a)^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02 \[ \int x^3 \arccos \left (a+b x^4\right ) \, dx=\frac {{\left (b x^{4} + a\right )} \arccos \left (b x^{4} + a\right ) - \sqrt {-b^{2} x^{8} - 2 \, a b x^{4} - a^{2} + 1}}{4 \, b} \]

[In]

integrate(x^3*arccos(b*x^4+a),x, algorithm="fricas")

[Out]

1/4*((b*x^4 + a)*arccos(b*x^4 + a) - sqrt(-b^2*x^8 - 2*a*b*x^4 - a^2 + 1))/b

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.30 \[ \int x^3 \arccos \left (a+b x^4\right ) \, dx=\begin {cases} \frac {a \operatorname {acos}{\left (a + b x^{4} \right )}}{4 b} + \frac {x^{4} \operatorname {acos}{\left (a + b x^{4} \right )}}{4} - \frac {\sqrt {- a^{2} - 2 a b x^{4} - b^{2} x^{8} + 1}}{4 b} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {acos}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*acos(b*x**4+a),x)

[Out]

Piecewise((a*acos(a + b*x**4)/(4*b) + x**4*acos(a + b*x**4)/4 - sqrt(-a**2 - 2*a*b*x**4 - b**2*x**8 + 1)/(4*b)
, Ne(b, 0)), (x**4*acos(a)/4, True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int x^3 \arccos \left (a+b x^4\right ) \, dx=\frac {{\left (b x^{4} + a\right )} \arccos \left (b x^{4} + a\right ) - \sqrt {-{\left (b x^{4} + a\right )}^{2} + 1}}{4 \, b} \]

[In]

integrate(x^3*arccos(b*x^4+a),x, algorithm="maxima")

[Out]

1/4*((b*x^4 + a)*arccos(b*x^4 + a) - sqrt(-(b*x^4 + a)^2 + 1))/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int x^3 \arccos \left (a+b x^4\right ) \, dx=\frac {{\left (b x^{4} + a\right )} \arccos \left (b x^{4} + a\right ) - \sqrt {-{\left (b x^{4} + a\right )}^{2} + 1}}{4 \, b} \]

[In]

integrate(x^3*arccos(b*x^4+a),x, algorithm="giac")

[Out]

1/4*((b*x^4 + a)*arccos(b*x^4 + a) - sqrt(-(b*x^4 + a)^2 + 1))/b

Mupad [B] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.11 \[ \int x^3 \arccos \left (a+b x^4\right ) \, dx=\frac {x^4\,\mathrm {acos}\left (b\,x^4+a\right )}{4}-\frac {\sqrt {-a^2-2\,a\,b\,x^4-b^2\,x^8+1}}{4\,b}-\frac {a\,\ln \left (\sqrt {-a^2-2\,a\,b\,x^4-b^2\,x^8+1}-\frac {b^2\,x^4+a\,b}{\sqrt {-b^2}}\right )}{4\,\sqrt {-b^2}} \]

[In]

int(x^3*acos(a + b*x^4),x)

[Out]

(x^4*acos(a + b*x^4))/4 - (1 - b^2*x^8 - 2*a*b*x^4 - a^2)^(1/2)/(4*b) - (a*log((1 - b^2*x^8 - 2*a*b*x^4 - a^2)
^(1/2) - (a*b + b^2*x^4)/(-b^2)^(1/2)))/(4*(-b^2)^(1/2))